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In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

For a positive-mass object in the inverse-square attractive field with potential zero at infinity, energy conservation gives $v^2(r)=2E/m+2GM/r$. Along the outgoing unbound branch, speed decreases toward the displayed limiting value. It is positive for a hyperbola and zero for a parabola. This is an infimum approached at infinite separation, not a finite-radius stopping point or the local escape velocity.

A steady outward radial flow with negligible pressure force conserves $v^2/2-GM/r$. If its speed at a reference radius equals the mechanical escape velocity, this constant is zero and $v\propto r^{-1/2}$. Conservation of mass gives $\rho \propto r^{-3/2}$. An adiabatically evolving ideal gas carried along that leading flow has $p\propto r^{-3\gamma /2}$ and $T\propto r^{-3(\gamma -1)/2}$.

Finite specific enthalpy changes the dynamical problem: the full conserved Bernoulli equation is $v^2/2+h-GM/r=B$. At the same boundary mechanical escape speed, $B=h_e$ rather than zero. A warm wind therefore needs its enthalpy and equation of state specified before the explicit ballistic velocity can be replaced by a consistent pressure-driven solution.

For equal-density spherical planetesimals, an impactor much smaller than target diameter $D$ causes a catastrophic planetesimal collision when its specific impact energy reaches $Q_D^{\ast }$. Neglecting gravitational focusing,When encounter speed is much smaller than the target's escape velocity, impact speed is approximately $v_{\mathrm{esc}}=D\sqrt{2\pi G\rho /3}$. For $Q_D^{\ast }=Q_b{D}^b$, the focusing-dominated threshold is $D_{\mathrm{cc}}=(3Q_b/\pi G\rho {)}^{1/3}{D}^{(1+b)/3}$. This assumes the disruption threshold uses impact energy per target mass and neglects energy-coupling efficiency variations.

For a sphere of radius $r_0$ and uniform mass density ${\rho }_0$, with no surrounding mass and gravitational potential zero at infinity, the spherical shell theorem gives $V_c^2=4\pi G{\rho }_0{r}^2/3$ inside. Matching to the exterior point-mass potential gives $\Phi (r)=-(2\pi G{\rho }_0/3)(3r_0^2-r^2)$. Since $v_e^2=-2\Phi$, the displayed relation follows for $0<r\le r_0$. Exterior shells leave the interior circular speed unchanged but deepen the potential and change the escape velocity; a uniform core alone is insufficient to determine escape.

A spherical halo with flat circular speed $v_c$ has $M(r)=v_c^{2}r/G$ and $\rho =v_c^2/(4\pi G{r}^2)$. If it is sharply truncated at $r_{\ast }$, the work per unit mass from $r<r_{\ast }$ to the boundary is $v_c^2\log (r_{\ast }/r)$ and from the boundary to infinity is $v_c^2$. Adding them gives the displayed escape velocity. A bound star's observed radial speed is only a lower bound on its total speed and on the escape speed, so one maximum radial measurement generally gives a lower bound on $r_{\ast }$ rather than an exact halo edge.

The escape velocity is the least initial speed that permits a freely coasting body to move arbitrarily far from an isolated attracting body. Despite its conventional name, it is a speed, not a unique velocity vector. For a fixed point mass $M$, or outside a body with spherical symmetry, set Newtonian gravitational potential energy zero at infinity. An exterior test particle of constant mass $m\ll M$ then hasHere $\mu =GM$ is the standard gravitational parameter, and $r$ is distance from the centre, not height above the surface. Set $r=R$ for surface escape velocity. Evaluate velocity in the body's rest inertial frame. Neglect drag, later rocket thrust, other bodies' gravity, and relativistic effects.^[ref]

Conservation of energy gives the threshold by requiring $E\ge 0$: at infinity the potential energy tends to zero and the kinetic energy cannot be negative. If $E=0$, the outward motion loses speed continuously toward zero. If $E>0$, the asymptotic speed is $v_{\infty }=\sqrt{2E/m}=\sqrt{v_0^2-v_{\mathrm{esc}}(r_0{)}^2}$. If $E<0$, infinity is inaccessible. For a radial outward launch, its outer turning point is $r_{\max }=-GMm/E$. For nonradial motion use angular momentum and the escape energy threshold to distinguish zero radial velocity from zero total speed.

Direction still determines the orbit and whether the body hits an obstacle. An initially inward velocity can cause collision even when $E\ge 0$. For an isolated spherical body, an exterior outward launch, or a tangential launch at or above escape velocity, has an outward escape branch without intersecting the body. A parabolic trajectory is the nonradial threshold path; a hyperbolic Kepler orbit has positive energy. The formula specifies their required local speed, not their orientation.^[ref]

For identical point masses of total mass $M$, let $W=-G{M}^2/R_g$ exclude self-interactions. The virial theorem gives $⟨v_{\mathrm{eq}}^2⟩=GM/R_g$. Since the sum of individual potential energies counts each pair twice, the mass-weighted mean frozen-potential escape velocity squared is $-4W/M$. A dissipation-free collapse from zero total energy instead has kinetic energy $-W$, hence mean squared speed $-2W/M$. These are three-dimensional speeds relative to the centre of mass; a coherent collapse speed is not a local random velocity dispersion.

Here inverse-square potential means the potential energy associated with an attractive inverse-square law for a central force: the potential energy itself varies as inverse distance. For $k>0$ and position $\mathbf{r}$ with $r=|\mathbf{r}|>0$,The gradient makes the distinction explicit: doubling $r$ halves the magnitude of potential energy but quarters the magnitude of force. The neighboring repulsive inverse-square potential instead has potential energy proportional to $r^{-2}$ and an inverse-cube force; the two naming conventions must be distinguished.

For Newton's law of universal gravitation, $k=GMm$, where $G$ is the gravitational constant. Treat the attracting mass $M$ as an isolated fixed point mass, or as a body with spherical symmetry and exterior test particle of constant mass $m\ll M$. The spherical shell theorem gives the same exterior Newtonian gravitational potential energy $U=-GMm/r$ without requiring uniform mass density. Here $r$ is measured from the centre, and the formula applies outside the body; it does not describe its interior. The Newtonian gravitational potential per unit mass is $\Phi =U/m=-GM/r$.^[ref]

The negative sign records binding relative to infinite separation, rather than a negative kinetic energy. Moving from $r_1$ to $r_2>r_1$ increases potential energy byBecause the force is a conservative force, this work depends on the endpoints, including for a curved path. Integrating $dV/dr=k/r^2$ with zero at infinity gives $V=-k/r$; adding a constant leaves the force unchanged but shifts the numerical escape energy threshold by that same constant. This is why the reference at infinity must accompany an escape velocity calculation.

For an exact synthetic comparison, stipulate the standard gravitational parameter $\mu =GM=8{\mathrm{m}}^3/{\mathrm{s}}^2$ and a $1\mathrm{kg}$ test particle. At $r=2\mathrm{m}$, $U=-4\mathrm{J}$ and the inward force has magnitude $2\mathrm{N}$; at $r=4\mathrm{m}$, $U=-2\mathrm{J}$ and the inward force has magnitude $0.5\mathrm{N}$. The increase is $2\mathrm{J}$ of potential energy. These stipulated parameters are an arithmetic example, not observations of a particular body. The escape-energy balance diagram uses the same parameters.

A lowered Maxwellian velocity distribution subtracts the value of a Maxwell-Boltzmann velocity distribution at the escape velocity and sets the result to zero beyond escape. This gives a nonnegative phase-space distribution function vanishing continuously at the boundary. If $A=n_0/(2\pi {\sigma }^2{)}^{3/2}$ and $x_e=v_e/(\sqrt{2}\sigma )$, its actual number density is $n=n_0[\mathrm{erf}{x}_e-2x_e{e}^{-x_e^2}/\sqrt{\pi }-4x_e^3{e}^{-x_e^2}/(3\sqrt{\pi })]$. Thus $n_0$ is a normalization parameter, not generally the integrated density. Integrating only speeds below $V<v_e$ gives the same bracket with the first two terms at $X=V/(\sqrt{2}\sigma )$ and the last term $-4X^3{e}^{-x_e^2}/(3\sqrt{\pi })$.

Lunar laser ranging gives a more discriminating orbit readout. An observatory directs short laser pulses at a retroreflector array on the Moon, and a detector records returning photons. Round-trip time, statistically combined over several returns, constrains distance; pressure and temperature accompany the readout. The fitted model includes Newtonian gravity, terrestrial and lunar rotation, geophysical effects, and corrections from general relativity. Comparing orbit determination residuals across stations and model choices checks consistency. The inspected study reports $1.8\mathrm{cm}$ weighted root mean square residuals for its 2005–2011 ranges, rather than for a bare isolated inverse-square potential. It tests the equivalence principle using the Earth and Moon's response to the Sun; it is not a direct escape velocity measurement. Uneven coverage near new and full Moon, radiation pressure, and thermal expansion of reflector mounts limit that inference.^[ref]

Equating the molecular root-mean-square speed to the surface escape velocity gives $3k_{B}T/m=2GM/R$, or $T=2mGM/(3k_{B}R)$. This compares speed scales; the high-speed tail can escape below that temperature, and molecular dissociation can invalidate the assumed species at high temperature.

The Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

In the stated gravity-dominated approximation, neglect the pressure force in the radial momentum equation. A steady outward stellar wind then obeysAt $R_e=R_{\mathrm{esc}}$, the escape velocity is $v_e=\sqrt{2G{M}_{\ast }/R_e}$, so $C=0$. The ballistic marginal-escape stellar wind hasThe outward flow decelerates and approaches zero speed at infinity; it is exactly at the escape threshold in this pressure-negligible model. An initial speed above that threshold gives a nonzero asymptotic speed.

Adiabaticity alone would instead conserve the full Bernoulli equation $v^2/2+h-G{M}_{\ast }/r=B$. If finite pressure is retained and $v_e$ equals the mechanical escape speed, then $B=h_e$, not zero. With $p=K{\rho }^{\gamma }$ and conservation of mass, the more general implicit solution isIt needs $h_e$ and $\gamma$ in addition to the escape-speed datum. The boxed explicit power law is the leading gravity-dominated result, not an exact solution for an arbitrary warm adiabatic wind.

Take the gravitational potential to vanish at infinity. At the position of star $i$, excluding its own point-mass field, the instantaneous test-particle escape velocity obeys $v_{\mathrm{esc},i}^2=-2{\Phi }_i$. The factor of one half in the total gravitational potential energy is important:Therefore the equal-mass average isThus the rms escape speed is twice the equilibrium rms speed. This is the virial mean escape speed relation, using the same three-dimensional averaging convention as the preceding part. It concerns escape in the instantaneous system potential, rather than a subsequent collective rearrangement of the cluster.

For an isolated smooth mass distribution with spherical symmetry, assemble each shell $dM$ at radius $r$ against the Newtonian gravitational potential of its interior mass. The spherical shell theorem givesAssume finite total mass, finite binding energy and no central point-mass self-energy, so $M(r{)}^2/r$ vanishes at both endpoints. Integration by parts then givesFor example, $M(r)\propto r^3$ near a regular center guarantees the central boundary condition, while finite total mass guarantees the one at infinity. A divergent central binding energy would fall outside this formula's finite-energy hypotheses.

For the discrete system use the centre of mass frame, let $N$ be the particle number, and write $M_{\mathrm{tot}}=Nm$. The printed $M$ in the particle-count factor and the subsequent $n$ must be understood consistently as $N$. DefineBy Newton's second law, differentiation givesEach pair contributesTherefore the exact dynamical virial theorem is $\ddot{I}/2=2K+W$. In a stationary statistical state, or on a bounded-motion time average for which the mean $\ddot{I}$ vanishes, it becomesThere is no external pressure, tidal field or boundary work in this isolated point-particle derivation. For a time-averaged system, the following $W$ and gravitational radius refer to that same averaged binding energy.

Using $W=-G(Nm{)}^2/R_g$ and $2K=Nm⟨v_{\mathrm{eq}}^2⟩$, virial equilibrium givesThis is the three-dimensional mean squared speed about the centre of mass. For an isotropic system each one-dimensional velocity dispersion squared is one third of it.

Set the Newtonian gravitational potential to zero at infinity and exclude each particle's own field. The local escape velocity in the instantaneous frozen potential obeys $v_{\mathrm{esc},i}^2=-2{\Phi }_i$, where ${\Phi }_i=-Gm{\sum }_{j\ne i}{r}_{ij}^{-1}$. Since every pair occurs twice, ${\sum }_{i}m{\Phi }_i=2W$. Averaging over the identical particles proves the gravitational rms escape and collapse speeds relationAn actual encounter in a changing potential can exchange energy, so this is the usual local escape-speed definition, not a claim about every many-body trajectory.

Finally, dissipation-free collapse from rest at infinite separation has total energy zero. Conservation of energy at the specified binding energy therefore gives $K_{\mathrm{coll}}=-W$, twice the equilibrium kinetic energy at the same $W$. HenceThis mean squared speed includes coherent infall; a local random velocity dispersion after subtracting the infall velocity need not obey that ratio.

Taking the scalar product of the Lorentz force equation with the velocity givesThe triple product vanishes because the cross product is perpendicular to its first factor. Thus the kinetic energy is constant, even if the magnetic field varies with position: magnetic forces do no work.

In cylindrical coordinates, the radial, angular and vertical equations areFor a nonstationary circular orbit with constant $r$ and $z$, the first equation gives $-m{\dot{\theta }}^2=qB(r)\dot{\theta }$. Hence its signed angular velocity is $\boxed{\dot{\theta }=-qB(r)/m}$. The sign gives the sense of circulation. The zero root describes a stationary transverse position, not a moving circular orbit. A constant vertical velocity can be superposed to produce a helix.

Define the mechanical angular momentum about the axis by $L=m{r}^2\dot{\theta }$. Multiplying the angular equation by $r$ givesFor the inverse-radius magnetic field this becomes $\dot{L}=-q{B}_0\dot{r}$. Integrating from an initial radius $r_0$ provesLet $c=q{B}_0/m$ and $\ell =(L_0+q{B}_0{r}_0)/m$, so that $L/m=\ell -cr$. Since $\dot{z}$ is constant and the total kinetic energy is constant, the transverse speed squaredis also constant. Substituting the angular momentum relation gives the required radial curve:Only the region $W(r)\ge 0$ is accessible. For $\ell \ne 0$, $W(r)\to -\infty$ as $r\to 0$, and its horizontal asymptote is $v_{\perp }^2-c^2$. Its derivative is $W^{\prime }(r)=2\ell (\ell -cr)/r^3$. Thus, when $\ell c>0$, it has a single maximum $W=v_{\perp }^2$ at $r=\ell /c$. The positive roots of $W$ are the turning points obtained from $r=\ell /(c\pm v_{\perp })$.

For nonzero $c$ and $\ell c>0$, write $r_c=|\ell /c|$. When $0<v_{\perp }<|c|$, two turning points occur atand the radius is confined between them. At $v_{\perp }=|c|$, there is only the finite lower turning point $r_c/2$; $W$ tends to zero from above and an outgoing particle reaches arbitrarily large radius. For $v_{\perp }>|c|$, only the lower turning point remains and the outgoing radial velocity approaches $\sqrt{v_{\perp }^2-c^2}$. The sketch shows these three cases and omits inaccessible negative values of ${\dot{r}}^2$ from the physical interpretation.

For completeness, if $\ell c<0$, $W$ rises monotonically from $-\infty$ to its asymptote; admissible motion then requires $v_{\perp }>|c|$ and is radially unbound after any inward branch turns. If $\ell =0$, $W=v_{\perp }^2-c^2$ is constant; equality gives a circular orbit rather than escape. Above threshold the outgoing branch escapes, whereas an incoming branch reaches the singular axis in finite time and cannot be continued through it without specifying a nonsingular core field. These cases explain why a borderline escape assertion needs an initial-direction qualification.

The transverse escape velocity threshold for radial escape in an inverse-radius axial magnetic field isFor a radial launch at $r_0$, $\ell =c{r}_0$, and equality permits marginal escape; for $\ell \ne 0$, larger transverse speed guarantees radial escape in forward time after any initial inward leg turns. The singular incoming $\ell =0$ branch is the exception described above. More general launches at equality can instead give the circular exception $\ell =0$. A vertical velocity adds to the total speed but cannot pay the transverse escape-energy requirement; if escape merely means unbounded $z$, any nonzero $\dot{z}$ already achieves that. If $q{B}_0=0$, transverse motion is force-free and there is no positive escape-speed threshold.

The stated relation follows for an isolated uniform sphere ending at $r_0$, with the gravitational potential zero at infinity. Uniform density in a core alone does not fix the escape velocity: exterior shells add a constant interior gravitational potential while leaving the interior force unchanged.

Under the isolated-sphere assumption, the enclosed mass and circular speed areThe boundary value is $\Phi (r_0)=-Gm(r_0)/r_0$. Integrating ${\Phi }^{\prime }=Gm(r)/r^2$ inwards givesThus the escape and circular speeds inside an isolated uniform sphere satisfyThe finite central limit is $v_e^2(0)=4\pi G{\rho }_0{r}_0^2$, although $V_c(0)=0$. Adding an exterior spherical shell of mass $m_s$ at radius $s>r_0$ raises $v_e^2$ by $2G{m}_s/s$ and is a direct counterexample to deducing the boxed relation from a uniform core alone.

For a fixed central gravitational source, the radial and transverse components of Newton's second law in polar coordinates areThe second equation gives $r^2\dot{\theta }=h$, the constant specific angular momentum. Take $h>0$ by choosing the orientation of $\theta$, and put $u=1/r$. Writing $u^{\prime }$ and $u^{\prime \prime }$ for derivatives with respect to $\theta$,Substituting into the radial equation and dividing by $-h^2{u}^2$ proves the Binet equationAt perihelion the velocity is tangential. The escape velocity is $v_0=\sqrt{2GM/r_0}$, so $h=r_0{v}_0$ and $h^2=2GM{r}_0$. Consequently $k=1/(2r_0)$. The initial conditions are $u(0)=1/r_0$ and $u^{\prime }(0)=0$. Solving the linear ordinary differential equation gives the parabolic Kepler orbitUsing angular momentum conservation again,On the outgoing branch $r=2r_0$ is first reached at $\theta =\pi /2$. Integrate from release at perihelion, using $w=\tan (\theta /2)$:More generally the same integral gives $t=(2r_0^2/h)(w+w^3/3)$, the Barker equation for this parabolic orbit.

The escape-speed scaling of wind mass loading makes the proposed trend dynamically plausible: the shallower Newtonian gravitational potential of a dwarf galaxy lets stellar feedback expel more gas per unit stellar mass. If coupled feedback supplies specific energy $e_{\ast }$, escape requires $\nu v_{\mathrm{esc}}^2/2\lesssim e_{\ast }$; fixed coupling suggests $\nu \propto v_{\mathrm{esc}}^{-2}$ for an energy-driven outflow. Fixed specific momentum $p_{\ast }$ instead gives $\nu v_{\mathrm{esc}}\lesssim p_{\ast }$, or $\nu \propto v_{\mathrm{esc}}^{-1}$ for a momentum-driven outflow.

Since escape velocity has $v_{\mathrm{esc}}^2\sim GM/R$, galaxies whose characteristic radius grows more slowly than their gravitating mass have increasing binding energy per unit mass with increasing mass. For $R\propto M^a$ with $a<1$, the energy-driven outflow estimate gives $\nu \propto M^{a-1}$. Equivalently, greater velocity dispersion in massive spheroids implies greater escape energy and more effective retention of metals.

For scale, one core-collapse supernova of $10^{51}\mathrm{erg}$ per $100M_{\odot }$ formed supplies about $5\times 10^{15}\mathrm{c}{\mathrm{m}}^2{\mathrm{s}}^{-2}$ before coupling losses. At ten percent coupling the energetic upper estimate $2e_{\ast }/v_{\mathrm{esc}}^2$ is about $110$ for $v_{\mathrm{esc}}=30\mathrm{km}{\mathrm{s}}^{-1}$, but about $0.4$ for $500\mathrm{km}{\mathrm{s}}^{-1}$. Efficient loss from shallow potentials and retention in deep potentials can span the required range. These are illustrative energetics rather than a unique prediction: cooling, entrainment, fallback and the actual initial mass function determine the escaping mass-loading factor.